求定积分∫(x+2)dx/根号2x+1.上限4,下限0.需过程.谢谢!

t=√(2x+1) 原积分式变为
1/2∫(t^2+3)dt，上限为3，下限为1
结果为22/3

=∫(1/2*(2x+1)^(1/2)+3/2*(2x+1)^(-1/2))dx 上限4，下限0.
=1/3*(2x+1)^(3/2)+3(2x+1)^(1/2)
=9+9-1/3-3
=44/3

∫[(x+2)/根号2x+1]dx
= ∫{[1/2(2x+1)+2/3]/根号2x+1}dx
=∫1/2(根号2x+1)dx+∫[3/2 /根号2x+1 ]dx
=1/6 *(2x+1)^(3/2)+3/2*(根号2x+1)
上限4,下限0.
得到 =22/3